Question

Four particles A,B,C and D each of mass m are kept at the corners of a square of side L. Now the particle D is taken to infinity by an external agent keeping the other particles fixed at their respective positions. The work done by the gravitational force acting on the particle during its movement is....?

Answer 1

Gravitational potential at point D:

V1=−GmL−GmL−Gm2√L     =−GmL( 2+12√)

Gravitational potential at infinity:

V2=0

Work done:

W=m(V2−V1)    =m[0+GmL(22√+12√)]    =Gm2L(22√+12√

Answer 2

Given that,

Mass of four particle A, B, C and is m.

Side of square = L

We need to calculate the diagonal distance

Using pythagorean theorem

$DB=\sqrt{(DC)^2+(CB)^2}$

Put the value into the formula

$DB=\sqrt{L^2+L^2}$

$DB=\sqrt{2}L$

We need to calculate the gravitational potential at point D

Using formula of gravitational potential

$U_{1}=-\dfrac{Gm^2}{r_{DA}}-\dfrac{Gm^2}{r_{DC}}-\dfrac{Gm^2}{r_{DB}}$

Put the value into the formula

$U_{1}=-\dfrac{Gm^2}{L}-\dfrac{Gm^2}{L}-\dfrac{Gm^2}{\sqrt{2}L}$

The gravitational potential is zero at infinity.

We need to calculate the work done by the gravitational force acting on the particle during its movement

Using formula of work done  

$W=U_{2}-U_{1}$

Put the value into the formula

$W=0-(-\dfrac{Gm^2}{L}-\dfrac{Gm^2}{L}-\dfrac{Gm^2}{\sqrt{2}L})$

$W=\dfrac{Gm^2(2\sqrt{2}+1)}{\sqrt{2}L}$

Hence, The work done by the gravitational force acting on the particle during its movement is $\dfrac{Gm^2(2\sqrt{2}+1)}{\sqrt{2}L}$